The correct options are
A a system of hyperbola
C y2=x(1+x)−1
Rewriting the given equation as
2xydydx−y2=1+x2
⇒2ydydx−1xy2=1x+x. Putting y2=u.
We have dudx−1xu=1x+x.
The I.F. of this differential equation is 1x, so
u1x=∫(1x2+1)dx=−1x+x+C
⇒y2=(x2−1)+Cx.
Since y(1)=1 so C=1,
hence y2=x(1+x)−1 which represents a system of hyperbola.